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Don saves $1 the first week, 2$ the second week, $3 the third week, and $4 the fourth week. If he continues in this way, how many weeks will it take before Don has saved at least $50?


Please show me the simplest and quickest way to solve this problem. EXPLAIN YOUR STEPS THOROUGHLY!!

Verified Answer :

superkoopasirf

This forms an arithmetic sequence, where the first term = 1, common difference = 1, and number of terms = n.

The formula for the sum of an arithmetic sequence is 

\(S_{n} = \frac{1}{2}n(2a + (n - 1)d)\)

Where a = the first term and d = the common difference.

We want the sum to be at least $50, so
\(50 \leq \frac{1}{2}n(2a + (n - 1)d)\)

Substituting in a and d
\(50 \leq \frac{1}{2}n(2(1) + (n - 1)(1))\)

Rearranging
\(50 \leq \frac{1}{2}n(2 + (n - 1))\)
\(50 \leq \frac{1}{2}n(1 + n)\)
\(100 \leq n(1 + n)\)
\(100 \leq n^2 + n\)
\(0 \leq n^2 + n - 100\)

\(\text{Let } n^2 + n -100 = 0\)
\(n = \frac{-1 \pm \sqrt{1-4(1)(-100)}}{2}\)
\(n = \frac{-1 \pm \sqrt{401}}{2}\)
\(n \geq 0 \implies n = \frac{-1 + \sqrt{401}}{2}\)
\(n \approx 9.51\)


So it will take 9.51 weeks, or 10 weeks to the nearest week.


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